poj_3080

来源：互联网发布：编程学习网编辑：程序博客网时间：2024/06/11 02:42

源链接：http://poj.org/problem?id=3080

题目大意就是找一个最长公共子串(长度要<3)，如果有多个，就输出字典序最小的那个。

因为数据较小，直接暴力即可。

暴力枚举某个字符串的长度为1的子串，长度为2的子串........，边枚举边记录符合条件的子串。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m;#define Mod 1000000007#define N 110#define M 10010char ch[15][65];const int len = 60;bool check(char *p){int len1 = strlen(p);int i;for(int s=1;s<n;s++){i=0;int flag = 0;while(i+len1<=len){int tmp = i;int j=0;while(p[j] == ch[s][tmp]&&j<len1){//注意加上j<len1j++;tmp++;}if(j == len1){flag = 1;break;}i++;}if(!flag)return false;}return true;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endif    int t;    sf(t);    while(t--){    sf(n);    for(int i=0;i<n;i++)    sfs(ch[i]);    char tmp[65];    char ans[65];    memset(tmp,'\0',sizeof tmp);    memset(ans,'\0',sizeof ans);    for(int i=1;i<=len;i++){    for(int j=0;j<len;j++){    int s=0;    if(j+i>len) continue;    for(int k=j;k<j+i;k++)    tmp[s++] =ch[0][k];    tmp[s] = '\0';    if(check(tmp)){    int len1 = strlen(ans);    int len2 = strlen(tmp);    if(len1<len2)    memcpy(ans,tmp,sizeof(tmp));    else if(len1 == len2 && strcmp(tmp,ans)<0)    memcpy(ans,tmp,sizeof(tmp));    }    }    }    int ans_len = strlen(ans);    if(ans_len<3)    printf("no significant commonalities\n");    else    printf("%s\n",ans);    }return 0;}

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